why is it triangular initially? does this bring some specific benefits?

doesn't that shape costs a ludicrous amount of propellant? or in the Sun-centered fixed frame of reference the actual orbit undergoes much smaller changes of direction than it would appear from the images shown? (and thus the total $\Delta V$ required is not as big)

4 Answers
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The triangular trajectory are hyperbolic orbits with respect to the comet and they'll (also, among other tasks also mentioned in the image you're attaching) serve to establish its mass. In essence scientists will be looking at how the comet's gravity changes these "straight" legs of the triangular Rosetta's quasi-orbit around comet 67P and estimate its density / mass more precisely.

The trajectory changing maneuvers aren't all that costly to the Rosetta spacecraft, and it was mentioned (during the briefing) that we're talking of delta-v of only a few meters per second during each of these maneuvers. So in a sense, since the comet's own gravity isn't all that great and the probe was already successfully injected into comet's own heliocentric orbit, they're not expensive to Rosetta's own propellants and would be more alike to satellite station-keeping maneuvers, like say the ones that satellites in halo or Lissajous orbits at Lagrange points would be performing, also on a fairly regular basis.

thanks. It would be nice to understand how such triangular orbit is better than a circular one to identify the density distribution (satellites that have being doing it for Earth have followed pretty much circular LEOs, afaik)
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FedericoAug 6 '14 at 13:33

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@Federico I do apologize for duplication, I just wanted to confirm that that is indeed Rosetta's planned trajectory even now when its bi-lobe nature was established. Why triangular? Well the clue is in the "hyperbolic" part, i.e. the probe has still some hyperbolic excess velocity w.r.t. the comet that it'll slowly reduce to enter more stable obit. Of course, since the comet's mass distribution isn't yet precisely established, trying to enter such orbit would just as well require constant corrections. Long triangular legs give scientists longer time to observe effects of comet's gravity on it.
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TildalWave♦Aug 6 '14 at 13:39

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@Federico: Rosetta is currently going faster than orbital velocity. To go into a circular trajectory, it would have to thrust continuously -- which probably wouldn't be any more expensive than its current triangular trajectory, but spending more time in free fall lets it make more accurate measurements. (I think.)
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Keith ThompsonAug 7 '14 at 2:29

delta-v of only a few meters per second during each of these maneuvers The delta-v of each maneuver was not even 1m/s (I've seen ~0.8m/s per maneuver).
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TonioElGringo2 days ago

@TonioElGringo If you have a source for this information, I'll include it in my answer (or, alternatively, feel free to edit it in), but that's what they said during the press briefing. I also thought it's a bit much for ε ~ -0.003 m²/s², but they were hyperbolic legs so I can't really calculate it. It does make sense that they would be as close to comet's C3 as possible though, since they would be measuring Rosetta's potential energy wrt the comet to refine its mass. So I tend to agree with you.
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TildalWave♦2 days ago

Adding to the existing answers: The triangular flight paths were needed to get enough images of the comet's surface, which are required for the low paths Rosetta flies today (and for dropping Philae close to the surface).

Rosetta's orbit propagator is an innovation insofar as it uses landmarks on the comet (among other things) to find the comet's and its own orientation, and to find a route to a given target landmark location. For this, the comet first had to be imaged in great detail, and that happened during the "triangle phase."

This has never been done before, and makes Rosetta a very exciting pilot project for ESA's future far-reaching space missions, which will all use this orbit propagator. It is already a breathtaking success if you consider the complexity the software has to manage.

Also, except for planned crash "landings," no spacecraft has ever flown by (or, as Rosetta does all the time now, has flown around and around) a comet in altitudes as low as 10 km. This requires a completely new auto-navigation system, and that one needs a map of the comet that cannot be acquired from earth (or from great distances in general).

(It is called "orbit propagator" despite the fact that Rosetta is not in an orbit because it is built upon other components that were core parts of "real" orbit propagators.)

It also has to do with the period. At 100km if Rosetta was just in a stable gravitational orbit due it would have a period of about 90 days, much less than the two weeks required. At 30km the gravitational period is about 2 weeks so this may explain why it is now in a circular orbit.

So for higher "orbits", the spacecraft flies in formation with the comet, more or less sharing the comet's heliocentric orbit. The spacecraft flies at just above its escape velocity from 67P, so a tiny thruster burn is enough to change course to the next leg of the triangle. If something happens to Rosetta (e.g. a debris collision), the spacecraft will continue to slowly move away from the comet, giving the team time to reestablish contact.

Why do those high orbits have a triangle shape: it's the shape with the fewest "corners": you only need 3 course corrections per orbit.

And why did ESA use higher orbits: directly after arrival, and now that the comet is approaching perihelion, they want to make sure the spacecraft doesn't get blasted by debris coming off the comet.